The area enclosed (in square units) by the curve y = x 4 - x 2 , the x -axis and the vertical lines passing through the two minimum points of the curve is

Step-by-step solution

Given that y = x 4 - x 2 For minimum points of curve: d d x x 4 - x 2 = 0 4 x 3 - 2 x = 0 2 x 2 x 2 - 1 = 0 ⇒ 2 x = 0 or 2 x 2 - 1 = 0 x = 0 or x = ± 1 2 So f is minimum at ± 1 2 The area = ∫ - 1 2 1 2 x 4 - x 2 d x   = 2 ∫ 0 1 2 x 4 - x 2 d x    ∵   f x     is even function = 2 x 5 5 - x 3 3 1 2 0 = 2 1 2 5 · 1 5 - 1 2 3 · 1 3 = 2 1 4 2 · 1 5 - 1 2 2 · 1 3 = 2 2 2 1 10 - 1 3 = 3 - 10 30 2 ∴ Area = 7 30 2